Page 414 - 35Linear Algebra

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we have
 
2
Lv =  −2  = 2v
4

so v ∈ E (2) . In general, we note the linearly independent vectors v (λ) with the
i
P i (λ)
same eigenvalue λ span an eigenspace since for any v = c v , we have
i i
X i (λ) X i (λ) X i (λ)
Lv = c Lv = c λv = λ c v = λv.
i i i
i i i

Hint for Review Problem 9

We are looking at the matrix M, and a sequence of vectors starting with

x(0)
v(0) = and defined recursively so that
y(0)

x(1) x(0)
v(1) = = M .
y(1) y(0)
We first examine the eigenvectors and eigenvalues of

3 2
M = .
2 3
We can find the eigenvalues and vectors by solving

det(M − λI) = 0

for λ.

3 − λ 2
det = 0
2 3 − λ
By computing the determinant and solving for λ we can find the eigenvalues λ =
1 and 5, and the corresponding eigenvectors. You should do the computations
to find these for yourself.
When we think about the question in part (b) which asks to find a vector
v(0) such that v(0) = v(1) = v(2) . . ., we must look for a vector that satisfies
v = Mv. What eigenvalue does this correspond to? If you found a v(0) with
this property would cv(0) for a scalar c also work? Remember that eigenvectors
have to be nonzero, so what if c = 0?
For part (c) if we tried an eigenvector would we have restrictions on what
the eigenvalue should be? Think about what it means to be pointed in the same
direction.

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